To make implementation and explanation a bit easier, let’s add a “border” to \(A\), that is, rows \(0\) and \(H+1\), similarly columns \(0\) and \(W+1\). They should be filled with a sentinel value that isn’t \(0\) or \(1\), including when “flipped” (e.g. \(2\)), and we also will stipulate that they will never count as isolated.

Now let’s define \(dp[i, a, b]\) as the minimum cost such that rows \(0\) to \(i\) inclusive are valid, and \(a\) and \(b\) are each \(0\) or \(1\), depending on whether rows \(i\) and \(i+1\) should be flipped, respectively. (If there are no valid states, \(dp[i, a, b] = \infty\).) We initialize the DP with \(dp[0, 0, 0] = 0\) and \(dp[0, 0, 1] = 1\), with the rest being \(\infty\). (as it makes no sense to “flip” the sentinel row)

We can transition from \(dp[i, a, b]\) to \(dp[i+1, b, c]\) because for each \(c \in \{0, 1\}\), we can check the \(i+1\)-st row for validity, since we know whether rows \(i\) through \(i+2\) should be flipped. This means we have \(2^3 = 8\) total cases/masks to check for each row, so the total time to run this DP is \(O(HW)\).

The final answer is \(\min(dp[H, 0, 0], dp[H, 1, 0])\). If this is \(\infty\), we should print \(-1\).